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cutiecomittee123
 one year ago
How do i solve this system of equations?
y=(x+1)^22
and y2x3
cutiecomittee123
 one year ago
How do i solve this system of equations? y=(x+1)^22 and y2x3

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whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.2is the second equation \[y=2x3\]?

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.2Well, let's assume that it is. \[y = (x+1)^22\]\[y=2x3\] We can solve this by substitution. Everywhere you see \(y\) in the first equation, replace it with the right hand side of the second equation (which is also equal to \(y\)). This gives us an equation all in terms of just the variable \(x\): \[2x3 = (x+1)^22\]Now you can expand that out: \[2x3 = (x+1)(x+1)2\]\[2x3=x^2+1x+1x+1*12\]\[2x3=x^2+2x1\]and get all the terms on one side and you have a quadratic to solve. You know how to do that?

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0The second equation is y2x=3

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.2Okay, can you solve \[y2x=3\] for \(y\)?

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.2Right. So now we do the same thing I did, except with the correct equation. \[y = (x+1)^2+2\]\[y = 2x+3\] \[2x+3 = (x+1)^2 + 2\] Expand both sides, then combine all the terms on one side and solve the quadratic.

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0wait it should be 3+2x=(x+1)^22

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.2doesn't \[2x+3 = 3+2x\]?

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.2sorry, I switched a sign, that's what you were pointing out, my bad!

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.2\[3+2x = (x+1)^22\]

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0yes but the original equation is y=(x+1)^22 and if we are plugging in y , and y=2x+3

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0Yeah thats what I got :))

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.2Glad one of us is paying close attention :)

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0so now i get 2x+3=x^2+1+2

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0thats after expanding the equation

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.2close... \[(x+1)^2 = (x+1)(x+1) = x*x + 1*x + 1*x + 1*1 = x^2 + 2x + 1\] so [2x+3 = x^2+2x+1  2\]\[2x+3 = x^2+2x1\] right?

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.2okay, get all the terms collected on one side, what do you get?

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.2Okay, and what are the solutions to that equation?

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.2That's a simple one to solve: \[x^24 = 0\]\[x^2 = 4\]Take the square root of both sides and you see \[x=\pm 2\] Have to be sure to include the negative square root or we won't get all of the answers! Now we need to find the values of \(y\) that go along with those values of \(x\). We can use either equation, so might as well use the simpler one: \[y=2x+3\] Plug in both values of \(x\) and get the corresponding values of \(y\).

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0okay just a sec

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0I get (2,7) and (2,1)

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.2and we can easily test them: \[2(2)+3 = (2+1)^22\]\[7 = 92\checkmark\] \[2(2)+3 = (2+1)^22\]\[1=12\checkmark\]

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.2So we can do this any time we can rearrange the stuff in one equation to give us something we can substitute into the other one.

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.2Here's a tip that doesn't really matter here so much, but does in problems that you will encounter: always try your answers out in the original equation and make sure they work! Whenever you have to square both sides of the equation, you can get what called "spurious" solutions which come out just the same way as the real ones do, but don't actually make the equations true when you plug the numbers in. Rather than remembering exactly what kinds of problems require you to do this, I prefer to simply always check my answers!
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